Which isomer would have the largest heat of combustion?

A) Propylcyclopropane
B) Ethylcyclobutane
C) Methylcyclopentane
D) Cyclohexane
E) Since they are all isomers, all would have the same heat of combustion.

Can anyone explain why?

If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion per bond than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.

It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.

For example:

Let's say a cyclopropane C−C bond is 10 J.
A cyclobutane C−C bond is 9.5 J.
A regular alkane bond is 8 J.

For propylcyclopropane:
Total energy is 54 J
For ethylcyclobutane:
Total energy is 54 J

The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.

How do you analyze this situation? Please help.


Compounds A-D all have the same molecular formula, $\ce{C6H12}$. We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation

$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$

Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).

This link provides the heats of combustion for some useful model compounds. Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained. The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.

Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".

  • $\begingroup$ Ron, in this case, you are using data to say that the TOTAL energy of the THREE C-C bonds cyclopropane is higher than the total energy of the FOUR C-C of cyclobutane. If we had only known that each C-C bonds in cyclopropane were higher in energy than the each one in cyclobutane, could we still be able to answer in the way you did? $\endgroup$ – yolo123 Sep 29 '14 at 2:03
  • $\begingroup$ Sorry for the nitty-gritty details, I'm just the kind of guy who obsesses about that kind of stuff! $\endgroup$ – yolo123 Sep 29 '14 at 2:03
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    $\begingroup$ yolo123, a) I'm using the data to say that a cyclopropane ring is always more strained than a cyclobutane ring, which is always more strained than a cyclopentane ring (by definition, a cyclohexane ring is strain free). b) Rather than view the data on a "per bond" basis, it is more commonly viewed on a "per CH2" basis - see the second to last column in the Table, and yes, just using this "per CH2" data we could have answered the question in the same way, e.g. how many CH2's are in each ring. $\endgroup$ – ron Sep 29 '14 at 2:11
  • $\begingroup$ Sorry for not being clear: I meant being able to answer this ONLY QUALITATIVELY. Without the data, just by knowing that strain/CH2 is higher in cyclopropane without any quantitative knowledge. After all, in an exam situation for which this question was made for, we do not have data. I guess we better know that the total strain is also higher in cyclopropane to answer this question. $\endgroup$ – yolo123 Sep 29 '14 at 2:25
  • $\begingroup$ Yes, just like we learn methyl, ethyl, propyl, we also learn that ring strain decreases from cyclopropane to cyclobutane (but they're close) to cyclopentane to cyclohexane. $\endgroup$ – ron Sep 29 '14 at 2:37

The products of combustion are the same so the compound with the highest potential energy (least stable) will have the highest enthalpy of combustion. All have the same number of carbons. The cyclopropane has the most ring strain and therefore the highest potential energy.

If you use metacognition you can see what's being asked. They can only write questions asking "Which has the most ring strain?" so many times. This question is supposed to involve more thinking but can be distilled down to "Which has the most ring strain?"

A good strategy in answering multiple choice questions is to determine what is the same and what is different about the answer choices. In this case the number of carbons is the same so the products are the same. What is different is the number of carbons in the rings and therefore the angle strain is different.

  • $\begingroup$ How can you know that 3 cyclopropyl bonds + another C-C alkly bond is more energetic than 4 cyclobutyl bonds? I know a cyclopropyl bond is more energetic than a cyclobutyl bond. $\endgroup$ – yolo123 Sep 29 '14 at 0:50
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    $\begingroup$ You can infer it from the intent of the question. There's no way it could be anything else and be a fair question. Chemistrywise look at the angles. 60 degrees is a big difference from 90 degrees. classes.yale.edu/chem220/STUDYAIDS/thermo/cycloalkanes/… sparknotes.com/chemistry/organic2/carbocycles/section1.rhtml $\endgroup$ – Brinn Belyea Sep 29 '14 at 0:53
  • $\begingroup$ Ok, so, there is nothing certain about the answer :S Are you sure there is no scientific way of proving this with the question without going to the "how to answer test questions realm" and looking up data? By the way, thanks a lot for the help. Really appreciate it. $\endgroup$ – yolo123 Sep 29 '14 at 0:57
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    $\begingroup$ The scientific way is to look at a heat of formation table. The thing is, the ring strain tables I found are close enough that I wonder if A really is the answer. It could be a bad question. $\endgroup$ – Brinn Belyea Sep 29 '14 at 0:59
  • $\begingroup$ Very confusing. $\endgroup$ – yolo123 Sep 29 '14 at 1:03

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